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BaLLatris [955]
3 years ago
7

Business intelligence is gained through industrial espionage and the gathering of this information often includes illegal or une

thical means.
A) True
B) False
Computers and Technology
1 answer:
Nataly [62]3 years ago
3 0

Answer:

A) True

Explanation:

Industrial espionage utilizes both illegal and unethical methods in gathering information about a corporate organization in order to get business intelligence. This involves stealing intellectual property and trade secrets to use them for a competitive advantage.  because Information about company's products, services, finances, sales, etc can be toold for  economic warfare

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Answer:

Check the explanation

Explanation:

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Kindly check the attached image below to get the step by step explanation to the above question.

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Write a do-while loop that asks the user to enter two numbers. The numbers should be added and the sum displayed. The user shoul
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The do-while loop for the given problem is shown below.  

do  

{  

  // user asked to enter two numbers  

  cout<<"Enter two numbers to be added."<<endl;  

  cin>>num1;  

  cin>>num2;  

  sum = num1 + num2;  

 

  // sum of the numbers is displayed  

  cout<<"Sum of the two numbers is "<<sum<<endl;  

  // user asked whether to perform the operation again  

  cout<<"Do you wish to continue (y/n) ?"<<endl;  

   cin>>choice;  

}while(choice != 'n');  

The variables to hold the two numbers and their sum are declared as float so that the program should work well for both integers and floating numbers.

float num1, num2, sum;

The char variable is taken since it holds only a single-character input from the user, 'y' or 'n'.

char choice;

The whole program is given below.

#include <iostream>

using namespace std;

int main() {

float num1, num2, sum;

char choice;  

do  

{  

  // user asked to enter two numbers  

  cout<<"Enter two numbers to be added."<<endl;  

  cin>>num1;  

  cin>>num2;  

  sum = num1 + num2;  

  // sum of the numbers is displayed  

  cout<<"Sum of the two numbers is "<<sum<<endl;  

  // user asked whether to perform the operation again  

  cout<<"Do you wish to continue (y/n) ?"<<endl;  

  cin>>choice;  

}while(choice != 'n');

cout<<"Quitting..."<<endl;

}

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